I have a huge expression that I have evidence that it can be simplified but don't know how During work on my thesis, I am working on formulating a solution to some problem and I came across a huge and very complicated formula.
I tried plugging in a lot of different values and I have evidence that it can be simplified.
The formula is as follows:
$n$ is an integer bigger than one 
$k$ is an integer which is smaller than $n$ and bigger than one.
$$\frac{1}{n}\sum_{i=1}^{n}{\sum_{r'=1}^{n}{ \sum_{c=0}^{i-1}\frac{{{r'-1}\choose{c}}{{n-r'}\choose{i-c-1}}}{{{n-1}\choose{i-1}}}\cdot\frac{  \sum_{r=n-k+1}^{n}{{{r-1}\choose{c}}{{n-r}\choose{i-c-1}}{}}}{ \sum_{r=c+1}^{n}{{{r-1}\choose{c}}{{n-r}\choose{i-c-1}}{}}}}}$$
As I mentioned, I plugged in lots of different values for $n$ and $k$ and the expression always evaluated $k$.  Is it true that it always is equal to $k$?
 A: Original goal
You want to show that $$\frac{1}{n}\sum_{i=1}^n \sum_{r'=1}^n \sum_{c=0}^{i-1} \frac{\binom{r'-1}{c} \binom{n-r'}{i-c-1}}{\binom{n-1}{i-1}} \frac{  \sum_{r=n-k+1}^n \binom{r-1}{c} \binom{n-r}{i-c-1} }{ \sum_{r=c+1}^{n}\binom{r-1}{c} \binom{n-r}{i-c-1} } = k$$
But some empirical investigation suggests that a stronger statement is true, so my revised goal is
$$\sum_{r'=1}^n \sum_{c=0}^{i-1} \binom{r'-1}{c} \binom{n-r'}{i-c-1} \frac{  \sum_{r=n-k+1}^n \binom{r-1}{c} \binom{n-r}{i-c-1} }{ \sum_{r=c+1}^{n}\binom{r-1}{c} \binom{n-r}{i-c-1} } = k \binom{n-1}{i-1}$$
Revised goal
If we start by reversing the order of the summations, we find that there's some symmetry which has been hidden:
$$\textrm{LHS} = \sum_{c=0}^{i-1} \frac{ \left[ \sum_{r=1}^n \binom{r-1}{c} \binom{n-r}{i-c-1} \right] \left[ \sum_{r=n-k+1}^n \binom{r-1}{c} \binom{n-r}{i-c-1} \right] }{ \sum_{r=c+1}^{n}\binom{r-1}{c} \binom{n-r}{i-c-1} } \\
= \sum_{c=0}^{i-1} \frac{ \sum_{r=1}^n \binom{r-1}{c} \binom{n-r}{i-c-1} }{ \sum_{r=c+1}^{n}\binom{r-1}{c} \binom{n-r}{i-c-1} } \sum_{r=n-k+1}^n \binom{r-1}{c} \binom{n-r}{i-c-1} \\$$
But when $r-1 < c$, we have $\binom{r-1}{c} = 0$, so the sum on top of the fraction in this new arrangement reduces to the sum on the bottom, and we have the much simpler
$$\textrm{LHS} = \sum_{c=0}^{i-1} \sum_{r=n-k+1}^n \binom{r-1}{c} \binom{n-r}{i-c-1} \\$$
Now let's substitute $s = n-r$:
$$\textrm{LHS} = \sum_{c=0}^{i-1} \sum_{s=0}^{k-1} \binom{n-s-1}{c} \binom{s}{i-c-1} \\$$
and it should be obvious that this can be identically equal to $k \binom{n-1}{i-1}$ iff $$s \ge 0 \implies \sum_{c=0}^{i-1} \binom{n-s-1}{c} \binom{s}{i-c-1} = \binom{n-1}{i-1}$$
But this is just a special case of Vandermonde's identity $$\sum_k \binom{r}{m+k} \binom{s}{n-k} = \binom{r+s}{m+n}$$
QED.
